Why is the statement "all vector space have a basis" is equivalent to the axiom of choice?

Question

I'm reading a section in an abstract algebra book, where it reviews vector spaces and suddenly comments that "all vector space have a basis" is equivalent to the axiom of choice...I haven't studied axioms of choice yet and after searching on the internet, I do not see why these two statements are equivalent...Could someone briefly explain to me? Thanks!

Answer 1

Note: It's worth pointing out that when we say "every vector space has a basis is equivalent to AC", we mean that these statements are equivalent over ZF (= "Zermelo-Fraenkel set theory without choice"). That is, the axiom system ZF can prove "AC iff every vector space has a basis."

The equivalence is not at all obvious! One implication is easy: using the axiom of choice to prove that every vector space has a basis. The other is the hard one, and was proved by Blass; see http://www.math.lsa.umich.edu/~ablass/bases-AC.pdf, which is self-contained.

Blass' construction actually proves that "every vector space has a basis" implies the axiom of multiple choice - that from any family of nonempty sets, we may find a corresponding family of finite subsets (so, not quite a choice function); over ZF this is equivalent to AC (this takes an argument, though, and in particular uses the axiom of foundation).

Very rough summary: start with a family $X_i$ of nonempty sets; without loss of generality, they're disjoint. Now look at the field $k(X)$ of rational functions over a field $k$ in the variables from $\bigcup X_i$. There is a particular subfield $K$ of $k(X)$ which Blass defines, and views $k(X)$ as a vector space over $K$. Blass then shows that a basis for $k(X)$ over $K$ yields a multiple choice function for the family $\{X_i\}$.

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The reason I don't give a better summary is that the full argument is really not reducible to a soundbite - if you want to understand it, you should read the details. There are many statements whose equivalence with AC has a "simple" picture; this is one of my favorite equivalences which is intricate!